Can I factor this?

ShawnGroskreutz · **Posted:** Sun, 15 Jul 2012 20:22:44 UTC

(1-x)^(gM/RL)

To

(1-x)^((gM/RL)-1)

Is this mathmatically correct?

Soroban · **Posted:** Sun, 15 Jul 2012 23:35:40 UTC

Hello, ShawnGroskreutz!

Quote:

$(1-x)^{\frac{gM}{RL}}\:\text{ to }\:(1-x)^{\frac{gM}{RL}-1}$

$\text{Is this mathmatically correct?}$

$\text{You just changed }\,(1-x)^4\,\text{ to }\,(1-x)^3$

$\text{What do }you\text{ think?}$

ShawnGroskreutz · **Posted:** Mon, 16 Jul 2012 00:46:14 UTC

Soroban,
Are you familiar with the Dervivation of the equation for Density Altitude based upon the three following
Equations from our 1976 standard atmosphere.

T=To-L*H

P=Po[1-(L*H/To)]^(g*M/R*L)

D=[P*M]/R*T*1000

I have had a very difficult time getting the correct equation: Which is

H=(To/L)[1-[(1000RToD/M*Po)]^[(L*R/g*M-L*R)]

I have been trying to get the derivation all day to no avail. Could you show a detailed derivation including how to get the exponential factor correct.

Shadow · **Posted:** Mon, 16 Jul 2012 02:56:44 UTC

That doesn't make any difference, you cannot subtract a power and expect it to be the same thing.

aswoods · **Posted:** Mon, 16 Jul 2012 09:47:29 UTC

You have confused the density altitude $h_\rho$ with the pressure altitude h_p

. The formula for $\tfrac{\rho}{\rho_0}$ is exactly the same as for $\tfrac{p}{p_0}$ except that the exponent is one less.

Denis · **Posted:** Mon, 16 Jul 2012 13:37:17 UTC

Shawn, go stand in the corner and repent :shock:

ShawnGroskreutz · **Posted:** Mon, 16 Jul 2012 15:50:37 UTC

$$ P=Po(1-\frac{L*H}{To})^\frac{g*M}{R*L}$

$$ \frac{P}{Po} = (1-\frac{L*H}{To})^\frac{g*M}{R*L}$

$$ (\frac{P}{Po})^\frac{R*L}{g*M} = 1-\frac{L*H}{To}$

$$ (\frac{P}{Po})^\frac{R*L}{g*M} = \frac{To-L*H}{To}$

$$ To\left(\frac{P}{Po}\right)^\frac{R*L}{g*M} = To - L*H$

$$ To\left(\frac{P}{Po}\right)^\frac{R*L}{g*M} - To = -L*H$

$$ \frac{-To}{L}\left(\left(\frac{P}{Po}\right)^\frac{R*L}{g*M} + 1 )\right= H$

$$ H = \frac{To}{L}\Left(1-\frac{P}{Po}^\frac{R*L}{g*M})\right$

$$ H = \frac{To}{L}\left(1- [\frac{D*R*T*1000}{Po*M}]^\frac{R*L}{g*M})\right$

At this point, the only difference between my result and the correct result is the exponential term. The correct equation is:

$$ H = \frac{To}{L}\left(1- \frac{D*R*T*1000}{Po*M}^\frac{R*L}{g*M-R*L})\right$

So, this is where I am stuck.

The three original equations were:

T = To -L*H

P = Po(1-(L*H/To))^[g*M/R*L}

D = P*M/R*T*1000

Please assist.

outermeasure · **Posted:** Mon, 16 Jul 2012 16:09:27 UTC

Substituting the first two equations into the third,
$\begin{aligned} D&=\frac{PM}{1000RT}\\ &=\frac{P_0\left(1-\frac{LH}{T_0}\right)^{\frac{gM}{RL}}M}{1000R(T_0-LH)}\\ &=\frac{P_0\left(1-\frac{LH}{T_0}\right)^{\frac{gM}{RL}}M}{1000RT_0(1-\frac{LH}{T_0})}\\ &=\frac{P_0\left(1-\frac{LH}{T_0}\right)^{\frac{gM}{RL}-1}M}{1000RT_0}\\ &=\frac{P_0\left(1-\frac{LH}{T_0}\right)^{\frac{gM-RL}{RL}}M}{1000RT_0} \end{aligned}$
and rearranging gives
$\displaystyle H = \frac{T_0}{L}\left[1- \left(\frac{1000DRT_0}{P_0M}\right)^\frac{RL}{gM-RL}\right].$

ShawnGroskreutz · **Posted:** Mon, 16 Jul 2012 17:42:09 UTC

That is it. Thank-you.

ShawnGroskreutz · **Posted:** Tue, 17 Jul 2012 17:34:56 UTC

$\displaystyle H = \frac{T_0}{L}\left[1- \left(\frac{1000DRT_0}{P_0M}\right)^\frac{RL}{gM-RL}\right].$ [/quote]

Hey, I need help evaluation the units of that particular equation. It seems to me that they don't all evaluate out to Kilometers.

Anyway:

To = 288.15 degrees kelvin
L = 6.5 degrees kelvin/KM
R = 8.31432 J/Mol degree K
g = 9.8065 m/sec^2
M = 28.9644 g/mol

Could someone show that the units all cancel out to Kilometers. I tried to evaluate the units and I am not getting a good result.

Shadow · **Posted:** Wed, 18 Jul 2012 04:56:37 UTC

ShawnGroskreutz wrote:

$\displaystyle H = \frac{T_0}{L}\left[1- \left(\frac{1000DRT_0}{P_0M}\right)^\frac{RL}{gM-RL}\right].$

Hey, I need help evaluation the units of that particular equation. It seems to me that they don't all evaluate out to Kilometers.

Anyway:

To = 288.15 degrees kelvin
L = 6.5 degrees kelvin/KM
R = 8.31432 J/Mol degree K
g = 9.8065 m/sec^2
M = 28.9644 g/mol

Could someone show that the units all cancel out to Kilometers. I tried to evaluate the units and I am not getting a good result.[/quote]

Just check that ${T_0\over L}$ gives km, the others are all pure numbers.

ShawnGroskreutz · **Posted:** Wed, 18 Jul 2012 16:05:08 UTC

Thanks, after knashing teeth for a while, I was able to prove that every unit cancels out, except for Km, and the rest of the equation is dimensionless.

Shadow · **Posted:** Wed, 18 Jul 2012 18:50:21 UTC

ShawnGroskreutz wrote:

Thanks, after knashing teeth for a while, I was able to prove that every unit cancels out, except for Km, and the rest of the equation is dimensionless.

I think you mean "gnashing", and I'm glad you figured it out.

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